Chapter 4.4, Problem 88E

Solution

a.

To graph: A scatter plot $(x,\sin x)$ for the special $17$ angles $x$ .

Given information:

Scatter plot for the function is $(x,\sin x)$ .

The intervals given are $-\pi \le x\le \pi$

Calculation:

The given points are,

  $(x,\sin x)$

Now to form a table data for $17$ special angles to get,

$x_1$	$\sin x_1$
$-\pi$	$0$
$-\frac{5\pi }{6}$	$-0.5$	$\frac{\pi }{6}$	$0.5$
$-\frac{3\pi }{4}$	$-0.70710678$	$\frac{\pi }{4}$	$0.70710678$
$-\frac{2\pi }{3}$	$-0.8660254$	$\frac{\pi }{3}$	$0.8660254$
$-\frac{\pi }{2}$	$-1$	$\frac{\pi }{2}$	$1$
$-\frac{\pi }{3}$	$-0.8660254$	$\frac{2\pi }{3}$	$0.8660254$
$-\frac{\pi }{4}$	$0.70710678$	$\frac{3\pi }{4}$	$0.70710678$
$-\frac{\pi }{6}$	$-0.5$	$\frac{5\pi }{6}$	$0.5$
$0$	0	$\pi$	0

Graph:

Plotting the $17$ points,

  

Interpretation:

A graph for the scatter plot $(x,\sin x)$ has been plotted for $17$ special angles.

b.

To find: The quadratic regression for the data.

The quadratic regression for the data is $y=0.024605x^4+0.0000015415x^3-0.441x^2-0.00001194x+0.97029$ .

Given information:

Scatter plot for the function is $(x,\sin x)$ .

The intervals given are $-\pi \le x\le \pi$

Calculation:

The given points are,

  $(x,\sin x)$

Now to form a table data for $17$ special angles to get,

$x_1$	$\sin x_1$
$-\pi$	$0$
$-\frac{5\pi }{6}$	$-0.5$	$\frac{\pi }{6}$	$0.5$
$-\frac{3\pi }{4}$	$-0.70710678$	$\frac{\pi }{4}$	$0.70710678$
$-\frac{2\pi }{3}$	$-0.8660254$	$\frac{\pi }{3}$	$0.8660254$
$-\frac{\pi }{2}$	$-1$	$\frac{\pi }{2}$	$1$
$-\frac{\pi }{3}$	$-0.8660254$	$\frac{2\pi }{3}$	$0.8660254$
$-\frac{\pi }{4}$	$0.70710678$	$\frac{3\pi }{4}$	$0.70710678$
$-\frac{\pi }{6}$	$-0.5$	$\frac{5\pi }{6}$	$0.5$
$0$	0	$\pi$	0

To determine the quadratic regression for the data

  $y=a{x}_1^3+b{x}_1^2+c{x}_1+d$

Taking the parameters from the above table,

  $\begin{array}{c}a=-0.087181\\ b=-2.91\times {10}^{-10}\\ c=0.82629\\ d=1.858\times {10}^{-9}\end{array}$

Substituting the above parameters in the expression to get,

  $y=-0.087181x^3-2.91\cdot {10}^{-10}{x}^2+0.82629x+1.858\cdot {10}^{-9}$

Therefore, the quadratic regression for the data is $y=-0.087181x^3-2.91\cdot {10}^{-10}{x}^2+0.82629x+1.858\cdot {10}^{-9}$ .

c.

To compare: The approximation to the sine function given by the cubic expression with the Taylor polynomial approximations.

The results are same when comparing the approximation to the sine function given by the cubic expression with the Taylor polynomial approximations.

Given information:

Scatter plot for the function is $(x,\sin x)$ .

The intervals given are $-\pi \le x\le \pi$

Calculation:

From part (b), the quadratic regression is given by,

  $y=-0.087181x^3-2.91\cdot {10}^{-10}{x}^2+0.82629x+1.858\cdot {10}^{-9}$

Taking 3 decimals approximately,

  $y=-0.087181x^3+0.82629x$

The Taylor polynomial expression is given by,

  $\begin{array}{c}\sin x\approx x-\frac{x^3}{3!}\\ \sin x\approx x-0.166x^3\end{array}$

Therefore, the cubic regression of the sine function and the Taylor’s approximation has the same results.